I had a object like this:
const myObject = {
0: 'FIRST',
10: 'SECOND',
20: 'THIRD',
}
I want to create a type with this object values, like this:
type AwesomeType = 'FIRST' | 'SECOND' | 'THIRD';
How to achieve this ?
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2 Answers
To get the variable (object) type you can use typeof operator. To prevent literal types widening you can use as const assertion:
const myObject = {
0: 'FIRST',
10: 'SECOND',
20: 'THIRD',
} as const;
type Values<T> = T[keyof T];
type AwesomeType = Values<typeof myObject>; // "FIRST" | "SECOND" | "THIRD"
5 Comments
Marcio Pamplona
Awesome.
as const solves the question. I've struggling to understand why typescript was returning only the value type and not the string const instead. Thanks man!
Subrato Pattanaik
AwesomeType is equivalent to
Record<string,string> in this case. I think he wants AwesomeType to be Record<string, "FIRST" | "SECOND" | "THIRD">. I added example in your code. Playgroud
Aleksey L.
@SubatoPatnaik, no the desired type is
type AwesomeType = 'FIRST' | 'SECOND' | 'THIRD'. Type which represents all possible object values, not the object itselfSubrato Pattanaik
@AlekseyL. Could you give an example with the playground? Btw I like your answer.
Aleksey L.
Do it this way
type AwesomeType = 'FIRST' | 'SECOND' | 'THIRD';
type MyType = Record<number, AwesomeType>
const myObject: MyType = {
0: 'FIRST',
10: 'SECOND',
20: 'THIRD',
}
see in this playground
